Two point charges $q_1$ and $q_2$ are $l$ distance apart. If one of the charges is doubled and the distance between them is halved,the magnitude of force becomes $n$ times,where $n$ is

The law governing the force between electric charges is known as

Two spheres having the same radius and mass are suspended by two strings of equal length from the same point in such a way that their surfaces touch each other. On depositing a charge of $2 \times 10^{-6} \ C$ on them,they repel each other such that in equilibrium,the angle between their strings becomes $60^{\circ}$. If the distance from the point of suspension to the center of the sphere is $10 \ cm$,find the mass of each sphere. $(k = 9 \times 10^9 \ SI, g = 10 \ ms^{-2})$. (in $kg$)

$A$ total charge $Q$ is broken into two parts $Q_1$ and $Q_2$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when:

The force between two charges $0.06\,m$ apart is $5\,N$. If each charge is moved towards the other by $0.01\,m$,then the force between them will become.........$N$

Two point charges $q_1$ and $q_2$ are $l$ distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is: